The area bounded by the parabola y^2 = x and | vedclass

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(A) Given curves are $y^2 = x$ and $2y = x$. To find the points of intersection,substitute $x = 2y$ into $y^2 = x$: $y^2 = 2y \Rightarrow y^2 - 2y = 0

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$4/3$

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(A) Given curves are $y^2 = x$ and $2y = x$. To find the points of intersection,substitute $x = 2y$ into $y^2 = x$: $y^2 = 2y \Rightarrow y^2 - 2y = 0 \Rightarrow y(y - 2) = 0$. Thus,the points of intersection are $y = 0$ and $y = 2$. The area $A$ bounded by the curves is given by the integral of the difference between the curves with respect to $y$: $A = \int_{0}^{2} (x_{line} - x_{parabola}) dy = \int_{0}^{2} (2y - y^2) dy$. Evaluating the integral: $A = [y^2 - \frac{y^3}{3}]_{0}^{2} = (2^2 - \frac{2^3}{3}) - (0 - 0) = 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3} 	ext{ sq. units}$.

The area bounded by the parabola $y^2 = x$ and the straight line $2y = x$ is

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